# Thread: Maclaurin series for Ln Z! (( Gamma Function ))

1. ## Maclaurin series for Ln Z! (( Gamma Function ))

2. Originally Posted by amro05
What is $F$? Is $\text{Ln}$ the principal branch of the logarithm?

3. Originally Posted by Drexel28
What is $F$? Is $\text{Ln}$ the principal branch of the logarithm?

4. Originally Posted by amro05
The 'factorial function' is defined as...

$\displaystyle z! = \int_{0}^{\infty} t^{z}\ e^{-t}\ dt = \Gamma(z+1)= z\ \Gamma(z)$ (1)

... and Your request [if I undestood correctly...] is to find the Mc Laurin series of $\ln z!$. Very well!... let's start from the well known 'inifinite product'...

$\displaystyle \frac{1}{\Gamma (z)} = z\ e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}}$ (2)

... where...

$\displaystyle \gamma = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n$ (3)

... is the so called 'Euler's constant'. From (1) and (2) we derive...

$\displaystyle \frac{1}{z!} = e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}}$ (4)

... and from (4)...

$\displaystyle \lambda(z)= \ln z! = -\gamma\ z - \sum_{n=1}^{\infty} \{\ln (1+\frac{z}{n}) - \frac{z}{n} \}$ (5)

Deriving both terms of (5) we obtain...

$\displaystyle \lambda^{'}(z) = -\gamma - \sum_{n=1}^{\infty} \{\frac{1}{n}\ (1+\frac{z}{n})^{-1} - \frac{1}{n} \}$ (6)

... so that is...

$\displaystyle \lambda^{'}(0) = -\gamma$ (7)

Deriving both terms of (6) we obtain...

$\displaystyle \lambda^{''}(z) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}\ (1+\frac{z}{n})^{-2}$ (8)

... so that is...

$\displaystyle \lambda^{''}(0) = \zeta (2)$ (9)

... where $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ is the so called 'Riemann zeta function'. Proceeding along the way we find for $n>1$ ...

$\displaystyle \lambda^{(n)}(0) = (-1)^{n}\ (n-1)!\ \zeta(n)$ (10)

... so that the requested McLaurin expansion is...

$\displaystyle \ln z! = -\gamma\ z + \sum_{n=2}^{\infty} (-1)^{n}\ \zeta(n)\ \frac{z^{n}}{n}$ (11)

The (11) converges for $|z|<1$...

Kind regards

$\chi$ $\sigma$

5. Many many thanks

all the best to you